当我试图从AJAX中的PHP文件接收JSON数据时,一切正常:
test.php的
<?php
header("Content-Type: application/json; charset=UTF-8");
$pdo = new PDO('mysql:host=localhost;dbname=database', 'root', '');
$stmt = $pdo->prepare("SELECT id, name FROM table");
$stmt->execute();
$stmt = $stmt->fetchAll(PDO::FETCH_CLASS);
echo json_encode($stmt);
?>
main.js
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
myObj = JSON.parse(this.responseText);
console.log(myObj);
document.getElementById("par").innerHTML = myObj[4].id + " " + myObj[4].tag1;
}
};
xmlhttp.open("GET", "http://localhost/project/app/models/Test.php", true);
xmlhttp.send();
但是当我试图从silex控制器接收JSON时:
Ajax.php
namespace AI\models;
use Silex\Application;
use Silex\Api\ControllerProviderInterface;
class Ajax implements ControllerProviderInterface
{
public function connect(Application $app){
$controllers = $app['controllers_factory'];
$controllers->get('/', 'AI\models\Ajax::home');
return $controllers;
}
public function home(Application $app){
header("Content-Type: application/json; charset=UTF-8");
$result = $app['db']->prepare("SELECT id, name FROM table");
$result->execute();
$result = $result->fetchAll(\PDO::FETCH_CLASS, \AI\models\Ajax::class);
echo json_encode($result);
return $app['twig']->render('ajax.twig');
}
}
main.js
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
myObj = JSON.parse(this.responseText);
console.log(myObj);
document.getElementById("par").innerHTML = myObj[4].id;
}
};
xmlhttp.open("GET", "http://localhost/project/app/models/Ajax.php", true);
xmlhttp.send();
我在控制台中收到错误:
Uncaught SyntaxError: Unexpected token < in JSON at position 0
at JSON.parse (<anonymous>)
at XMLHttpRequest.xmlhttp.onreadystatechange (main.js:4)
xmlhttp.onreadystatechange @ main.js:4
当我尝试从类中接收JSON时,我不知道为什么会出现问题。
答案 0 :(得分:0)
如果您要返回json,则应使用JsonResponse Object。你也绝对不想像现在那样渲染模板,你只需要json数据,为什么要渲染模板?
您可以按照之前关联的文档进行操作,但幸运的是Silex has a helper method for returning json data并且您甚至不需要使用JsonResponse(直接):
<?php
public function home(Application $app){
$result = $app['db']->prepare("SELECT id, name FROM table");
$result->execute();
$result = $result->fetchAll(\PDO::FETCH_CLASS, \AI\models\Ajax::class);
return $app->json($result);
}